Problem: $ F = \left[\begin{array}{rr}-2 & 1 \\ 3 & 0\end{array}\right]$ $ D = \left[\begin{array}{rrr}3 & -1 & 3 \\ -2 & -2 & -2\end{array}\right]$ What is $ F D$ ?
Answer: Because $ F$ has dimensions $(2\times2)$ and $ D$ has dimensions $(2\times3)$ , the answer matrix will have dimensions $(2\times3)$ $ F D = \left[\begin{array}{rr}{-2} & {1} \\ {3} & {0}\end{array}\right] \left[\begin{array}{rrr}{3} & \color{#DF0030}{-1} & \color{#9D38BD}{3} \\ {-2} & \color{#DF0030}{-2} & \color{#9D38BD}{-2}\end{array}\right] = \left[\begin{array}{rrr}? & ? & ? \\ ? & ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rrr}{-2}\cdot{3}+{1}\cdot{-2} & ? & ? \\ ? & ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rrr}{-2}\cdot{3}+{1}\cdot{-2} & ? & ? \\ {3}\cdot{3}+{0}\cdot{-2} & ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rrr}{-2}\cdot{3}+{1}\cdot{-2} & {-2}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{-2} & ? \\ {3}\cdot{3}+{0}\cdot{-2} & ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rrr}{-2}\cdot{3}+{1}\cdot{-2} & {-2}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{-2} & {-2}\cdot\color{#9D38BD}{3}+{1}\cdot\color{#9D38BD}{-2} \\ {3}\cdot{3}+{0}\cdot{-2} & {3}\cdot\color{#DF0030}{-1}+{0}\cdot\color{#DF0030}{-2} & {3}\cdot\color{#9D38BD}{3}+{0}\cdot\color{#9D38BD}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rrr}-8 & 0 & -8 \\ 9 & -3 & 9\end{array}\right] $